Consider the following fragment from chapter IV in Takesaki's book "Theory of operator algebra I":
Why is the boxed line true? Takesaki argues that
$$\theta_0: \mathscr{M}_1\otimes_{\operatorname{min}} \mathscr{M}_2 \to \mathscr{N}_1\otimes_{\operatorname{min}} \mathscr{N}_2: m_1 \otimes m_2 \mapsto \theta_1(m_1)\otimes \theta_2(m_2)$$
is $\sigma$-weakly continuous, but why does this ensure that the map $\theta_0$ extends to a $\sigma$-weakly continuous map on the von Neumann tensor products? I guess some general fact about topological vector spaces is invoked here.
Best Answer
The general setup in topological spaces would be the following: given topological spaces $X,Y$ and a dense $X_0\subseteq X$ with $T:X_0\rightarrow Y$ continuous (for the subspace topology on $X_0$) we can extend $T$ by continuity to $X$. This is of course not true (exercise to the reader to find a counter-example for e.g. $X_0=(0,1), X=[0,1]$).
A true statement can be fashioned by using the theory of Uniform Spaces, Uniform Continuity and Completeness. I believe that then a general statement could be formulated and proved for topological vector spaces.
However, in this case, I think there are some missing details in Takesaki's proof, and I would prefer to give a different approach.
The way I would think about this problem is to use some Banach space theory. The general setup is to let $E,F$ be Banach spaces, let $E_0\subseteq E^*$ be a weak$^*$-dense subspace, and let $T_0:E_0\rightarrow F^*$ be bounded, linear, and weak$^*$-continuous. We claim we can extend $T_0$ to $T:E^*\rightarrow F^*$ a weak$^*$-continuous bounded linear map.
In fact, Takesaki proves something different. We can consider the Banach space adjoint (here I follow Takesaki's notation) ${}^t T_0:F^{**}\rightarrow E_0^*$, consider $F$ as a subspace of $F^{**}$ and so ask about ${}^t T_0(F) \subseteq E_0^*$.
Also each member of $E$ induces a functional on $E_0$, and by Hahn-Banach, weak$^*$-density of $E_0$ in $E^*$ implies that the resulting map $E\rightarrow E_0^*$ is at least injective. Takesaki shows that ${}^t T_0(F) \subseteq E$. So there is a linear map $S:F\rightarrow E$ making the following commute: $$ \require{AMScd}\begin{CD} F^{**} @>{{}^t T_0}>> E_0^* \\ @AAA @AAA \\ F @>{S}>> E \end{CD} $$ Explicitly, given $\omega\in F$ and $x\in E_0\subseteq E^*$, $$ \langle x, S(\omega) \rangle_{E^*,E} = \langle {}^t T_0(\omega), x \rangle_{E_0^*, E_0} = \langle T_0(x), \omega \rangle_{F^*,F}. $$ In general, I see no reason why $S$ need be bounded. Kaplansky density comes to the rescue, because $M_1 \otimes_{\min} M_2$ is a weak$^*$-dense $C^*$-subalgebra of $M_1 \overline\otimes M_2$, and so the unit ball is dense. Abstractly, this means we may add the additional hypothesis that the unit ball of $E_0$ is weak$^*$-dense in the unit ball of $E^*$. Equivalently (by Hahn-Banach again) $E_0$ norms $E$. So \begin{align*} \|S(\omega)\| &= \sup \{ |\langle x,S(\omega)\rangle| : x\in E_0, \|x\|\leq 1 \} \\ &= \sup \{ |\langle T_0(x), \omega \rangle| : x\in E_0, \|x\|\leq 1 \} \\ &\leq \|T_0\| \|\omega\|. \end{align*} Thus $S$ is bounded, with $\|S\| \leq \|T_0\|$.
Now set $T = {}^t S_0 : E^*\rightarrow F^*$ and it's an easy exercise to show that $T$ extends $T_0$; clearly $T$ is weak$^*$-continuous as it's a Banach space adjoint.